|
1
|
Hi, let $k/\mathbb{Q}$ be a number field. Assume that $u$ is an algebraic integer such that all $k$-conjugates have modulus $1$. Is $u$ a root of $1$ ? If $k=\mathbb{Q}$, the answer is YES (this is Kronecker's theorem). I am pretty sure that this result is false if $k$ is an arbitrary number field, but I don't see any obvious counter-example. Any suggestion ? Thanks! |
||||||
|
|
4
|
The answer depends on the number field $k$. Of course, it cannot hold for all fields $k$, for if $u$ is an algebraic integer of modulus $1$ which is not a root of unity (there are plenty of them, see e.g. this MO-link), then set $k=\mathbb Q(u)$, so $u$ is the only $k$-conjugate of $u$. |
||
|
|
